If $x \barwedge y = 2x^{2}-y^{2}$ and $x \odot y = x^{2}-3y^{2}$, find $0 \barwedge (3 \odot 1)$.
First, find $3 \odot 1$ $ 3 \odot 1 = 3^{2}-3(1^{2})$ $ \hphantom{3 \odot 1} = 6$ Now, find $0 \barwedge 6$ $ 0 \barwedge 6 = 2(0^{2})-6^{2}$ $ \hphantom{0 \barwedge 6} = -36$.